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kipiarov [429]
2 years ago
5

Three pendulums with strings of the same length and bobs of the same mass are pulled out to angles θ1, θ2, and θ3, respectively,

and released. The approximation sin θ = θ holds for all three angles, with θ3 > θ2 > θ1. How do the angular frequencies of the three pendulums compare?
Physics
1 answer:
seropon [69]2 years ago
6 0

Answer:

The angular frequencies of all the 3 pendulums shall be same.

Explanation:

The time period of a simple pendulum with the approximation sin(\theta )\approx \thetais given by:

T=2\pi\sqrt{\frac{l}{g}}

The angular frequency \omega is given by

\omega =\frac{2\pi }{T}\\\\\omega =\frac{2\pi}{2\pi \sqrt{\frac{l}{g}}}\\\\\therefore \omega =\sqrt{\frac{g}{l}}

As we can see that the angular frequency is independent on the initial angle (valid strictly for small angle approximations) we conclude that the angular frequencies of the 3 pendulums are the same.

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When a surface is submerged in a fluid, the resultant pressure force on the body acts in what manner? Assume no shear forces are
Alex17521 [72]

Answer:

Perpendicular to the surface of the body.

Explanation:

When a surface is submerged in a fluid the resultant pressure force on the body acts perpendicular to the surface of the body. This is because fluids cannot withstand nor exert sideways forces. One could obtain this analytically considering that the origin of this force comes from the movement of the fluid molecules.

5 0
2 years ago
a 1.50*10^-5 C charge feels a 2.89*10^-3 N force when it moves 288m/s perpendicular (90) deg to a magnetic field. how strong is
Sunny_sXe [5.5K]
6.68, -1
Explanation: correct for acellus
5 0
2 years ago
Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Which student’s measureme
Lisa [10]

On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2 .</u>

<u>Explanation:</u>

Here, we have Four students measured the acceleration of gravity. The accepted value for their location is 9.78m/s2. Let's calculate which student’s measurement has the largest percent error :

<u>A. Student 4: 9.61 m/s2 </u>

Percentage of error  = \frac{9.78-9.61}{9.78} (100) = 1.738% %.

<u>B. Student 3: 9.88 m/s2 </u>

Percentage of error  = \frac{9.88-9.78}{9.78} (100) = 1.022% %.

<u>C. Student 2: 9.79 m/s2 </u>

Percentage of error  = \frac{9.79-9.78}{9.78} (100) = 0.1022%% .

<u>D. Student 1: 9.78 m/s2</u>

Percentage of error  = \frac{9.78-9.61}{9.78} (100) = 0%% .

On comparing values , we see that student which has the largest percent error is <u>A. Student 4: 9.61 m/s2 .</u>

4 0
2 years ago
A car experiences a centripetal acceleration of 4.4 m/s ^2 as ur rounds a corner with a speed of 15 m/s. What is the radius of t
damaskus [11]
The calculation of the centripetal acceleration of an object following a circular path is based on the equation,

                  a = v² / r

where a is the acceleration, v is the velocity, and r is the radius.

Substituting the known values from the given above,

             4.4 m/s² = (15 m/s)² / r

The value of r from the equation is 51.14 m.

Answer: 51.14 m
3 0
2 years ago
A wheel completes 5.6 revolutions in 8 seconds.
nevsk [136]

Answer:

86.15\pi rad/min

Explanation: Angular velocity is the number of revolutions made per unit time.

We convert the number of revolutions to radians and the time given in seconds to minutes,

Given;

1rev=2\pi rad\\therefore\\5.6rev=5.6*2\pi rad\\= 11.2\pi rad

Also,

60s = 1 min

hence

8s=\frac{8}{60}min\\=0.13min

We now divide the number of revolution in radians by the time in minutes.

\omega =\frac{11.2\pi}{0.13min}\\\omega=86.15\pi rad/min

5 0
2 years ago
Read 2 more answers
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